If $f(x)=mlogx+nx^2 + x $ has extreme values at x= 1 and x = -2, then the values of m and n respectively are :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $-2, \frac{1}{2}$
$\frac{d(f(x))}{dy}=\frac{d(m\log x+nx^2+x)}{dy}$
$=\frac{m}{x}+2nx+1$
$f'(1)=m+2n+1=0$
$⇒m+2n=-1$ ....(1)
$f'(-2)=\frac{-m}{2}-4n+1=0$
$-m-8n=-2$
$m+8n=2$ .....(2)
Eq. (2) - 4 × Eq. (1)
$-3m=6$
$m=-\frac{1}{2},n=-2$